Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $x = \dfrac{4y - 40}{y^2 - 8y - 20} \times \dfrac{-3y^2 + 3y + 18}{y + 2} $
Solution: First factor out any common factors. $x = \dfrac{4(y - 10)}{y^2 - 8y - 20} \times \dfrac{-3(y^2 - y - 6)}{y + 2} $ Then factor the quadratic expressions. $x = \dfrac {4(y - 10)} {(y + 2)(y - 10)} \times \dfrac {-3(y + 2)(y - 3)} {y + 2} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac {4(y - 10) \times -3(y + 2)(y - 3) } { (y + 2)(y - 10) \times (y + 2)} $ $x = \dfrac {-12(y + 2)(y - 3)(y - 10)} {(y + 2)(y - 10)(y + 2)} $ Notice that $(y + 2)$ and $(y - 10)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac {-12\cancel{(y + 2)}(y - 3)(y - 10)} {\cancel{(y + 2)}(y - 10)(y + 2)} $ We are dividing by $y + 2$ , so $y + 2 \neq 0$ Therefore, $y \neq -2$ $x = \dfrac {-12\cancel{(y + 2)}(y - 3)\cancel{(y - 10)}} {\cancel{(y + 2)}\cancel{(y - 10)}(y + 2)} $ We are dividing by $y - 10$ , so $y - 10 \neq 0$ Therefore, $y \neq 10$ $x = \dfrac {-12(y - 3)} {y + 2} $ $ x = \dfrac{-12(y - 3)}{y + 2}; y \neq -2; y \neq 10 $